3.2.48 \(\int \frac {1}{x^5 \sqrt {b x^2+c x^4}} \, dx\)

Optimal. Leaf size=80 \[ -\frac {8 c^2 \sqrt {b x^2+c x^4}}{15 b^3 x^2}+\frac {4 c \sqrt {b x^2+c x^4}}{15 b^2 x^4}-\frac {\sqrt {b x^2+c x^4}}{5 b x^6} \]

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Rubi [A]  time = 0.13, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2016, 2014} \begin {gather*} -\frac {8 c^2 \sqrt {b x^2+c x^4}}{15 b^3 x^2}+\frac {4 c \sqrt {b x^2+c x^4}}{15 b^2 x^4}-\frac {\sqrt {b x^2+c x^4}}{5 b x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*Sqrt[b*x^2 + c*x^4]),x]

[Out]

-Sqrt[b*x^2 + c*x^4]/(5*b*x^6) + (4*c*Sqrt[b*x^2 + c*x^4])/(15*b^2*x^4) - (8*c^2*Sqrt[b*x^2 + c*x^4])/(15*b^3*
x^2)

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps

\begin {align*} \int \frac {1}{x^5 \sqrt {b x^2+c x^4}} \, dx &=-\frac {\sqrt {b x^2+c x^4}}{5 b x^6}-\frac {(4 c) \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx}{5 b}\\ &=-\frac {\sqrt {b x^2+c x^4}}{5 b x^6}+\frac {4 c \sqrt {b x^2+c x^4}}{15 b^2 x^4}+\frac {\left (8 c^2\right ) \int \frac {1}{x \sqrt {b x^2+c x^4}} \, dx}{15 b^2}\\ &=-\frac {\sqrt {b x^2+c x^4}}{5 b x^6}+\frac {4 c \sqrt {b x^2+c x^4}}{15 b^2 x^4}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{15 b^3 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 46, normalized size = 0.58 \begin {gather*} -\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (3 b^2-4 b c x^2+8 c^2 x^4\right )}{15 b^3 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*Sqrt[b*x^2 + c*x^4]),x]

[Out]

-1/15*(Sqrt[x^2*(b + c*x^2)]*(3*b^2 - 4*b*c*x^2 + 8*c^2*x^4))/(b^3*x^6)

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IntegrateAlgebraic [A]  time = 0.16, size = 46, normalized size = 0.58 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-3 b^2+4 b c x^2-8 c^2 x^4\right )}{15 b^3 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^5*Sqrt[b*x^2 + c*x^4]),x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-3*b^2 + 4*b*c*x^2 - 8*c^2*x^4))/(15*b^3*x^6)

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fricas [A]  time = 1.09, size = 42, normalized size = 0.52 \begin {gather*} -\frac {{\left (8 \, c^{2} x^{4} - 4 \, b c x^{2} + 3 \, b^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, b^{3} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

-1/15*(8*c^2*x^4 - 4*b*c*x^2 + 3*b^2)*sqrt(c*x^4 + b*x^2)/(b^3*x^6)

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giac [A]  time = 0.21, size = 90, normalized size = 1.12 \begin {gather*} \frac {20 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{2} c + 15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} b \sqrt {c} + 3 \, b^{2}}{15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

1/15*(20*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^2*c + 15*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))*b*sqrt(c) + 3*b^2)/(
sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^5

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maple [A]  time = 0.00, size = 50, normalized size = 0.62 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (8 c^{2} x^{4}-4 b c \,x^{2}+3 b^{2}\right )}{15 \sqrt {c \,x^{4}+b \,x^{2}}\, b^{3} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(c*x^4+b*x^2)^(1/2),x)

[Out]

-1/15*(c*x^2+b)*(8*c^2*x^4-4*b*c*x^2+3*b^2)/x^4/b^3/(c*x^4+b*x^2)^(1/2)

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maxima [A]  time = 1.43, size = 68, normalized size = 0.85 \begin {gather*} -\frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{15 \, b^{3} x^{2}} + \frac {4 \, \sqrt {c x^{4} + b x^{2}} c}{15 \, b^{2} x^{4}} - \frac {\sqrt {c x^{4} + b x^{2}}}{5 \, b x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

-8/15*sqrt(c*x^4 + b*x^2)*c^2/(b^3*x^2) + 4/15*sqrt(c*x^4 + b*x^2)*c/(b^2*x^4) - 1/5*sqrt(c*x^4 + b*x^2)/(b*x^
6)

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mupad [B]  time = 4.33, size = 42, normalized size = 0.52 \begin {gather*} -\frac {\sqrt {c\,x^4+b\,x^2}\,\left (3\,b^2-4\,b\,c\,x^2+8\,c^2\,x^4\right )}{15\,b^3\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(b*x^2 + c*x^4)^(1/2)),x)

[Out]

-((b*x^2 + c*x^4)^(1/2)*(3*b^2 + 8*c^2*x^4 - 4*b*c*x^2))/(15*b^3*x^6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{5} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(1/(x**5*sqrt(x**2*(b + c*x**2))), x)

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